Difference between revisions of "2014 AMC 12A Problems/Problem 25"
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==Solution 2== | ==Solution 2== | ||
− | + | The axis of <math>P</math> is inclined at an angle <math>\theta</math> relative to the coordinate axis, where <math>\tan\theta = \tfrac 34</math>. We rotate the coordinate axis by angle <math>\theta</math> anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let <math>(\widetilde{x}, \widetilde{y})</math> be the coordinates in the rotated system. Then <math>(x,y)</math> and <math>(\widetilde{x}, \widetilde{y})</math> are related by | |
+ | <cmath>\begin{align} | ||
+ | \nonumber x = \widetilde{x}\cos\theta -\widetilde{y}\sin\theta &= \tfrac 45 \widetilde{x} - \tfrac 35 \widetilde{y}, \\ | ||
+ | y = \widetilde{x}\sin\theta +\widetilde{y}\cos\theta &= \tfrac 35 \widetilde{x} + \tfrac 45 \widetilde{y} | ||
+ | \end{align}</cmath> | ||
+ | In the rotated coordinate system, the parabola has focus at <math>(0,0)</math> and the two pints on it are at <math>(5,0)</math> and <math>(-5,0)</math>. Therefore, the directrix is <math>\widetilde{y}=\pm 5</math>; we can, WLOG, choose <math>\widetilde{y}=-5</math>. For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is | ||
+ | <cmath>\begin{align}\tag{2} | ||
+ | \widetilde{x}^2+\widetilde{y}^2 = (\widetilde{y}+5)^2 \qquad &\Longrightarrow\qquad \widetilde{y} = \tfrac{1}{10}(\widetilde{x}^2-25) | ||
+ | \end{align}</cmath> | ||
+ | From <math>(1)</math> we have <math>|4x+3y|=5\widetilde{x}</math>, so we need <math>|\widetilde{x}|<200</math>. Substituting <math>(2)</math> in <math>(1)</math>, we get | ||
+ | <cmath>\begin{align*} | ||
+ | 50x &= 40 \widetilde{x} - 3 \widetilde{x}^2 + 75, \\ | ||
+ | 50y &= 30 \widetilde{x} + 4 \widetilde{x}^2 - 100 | ||
+ | \end{align*}</cmath> | ||
+ | For <math>x</math> to be an integer <math>\widetilde{x}</math> must be a multiple of 5; setting <math>\widetilde{x}=5a</math> we get | ||
+ | <cmath>2x = 8a - 3 a^2 + 3</cmath> | ||
+ | Now we need <math>a</math> to be odd, i.e. <math>\widetilde{x}=5a</math> is an odd multiple of <math>5</math>, in which case we get <math>y = 3 a + 2 a^2 - 2</math>, which is also an integer. The values that satisfy the given conditions correspond to <math>\widetilde{x}= \{\pm 5\cdot (2k-1)\mid k = 0, 1, \ldots , 19 \}={-195, -185, -175, ..., 195}</math>, and there are <math>\boxed{\textbf{(B)} \ 40}</math> such numbers. | ||
(Solution by Shaddoll) | (Solution by Shaddoll) |
Revision as of 17:44, 22 November 2021
Problem
The parabola has focus and goes through the points and . For how many points with integer coordinates is it true that ?
Solution
The parabola is symmetric through , and the common distance is , so the directrix is the line through and , which is the line Using the point-line distance formula, the parabola is the locus which rearranges to .
Let , . Put to obtain and accordingly we find by solving the system that and .
One can show that the values of that make an integer pair are precisely odd integers . For this is , so values work and the answer is .
(Solution by C-273)
Solution 2
The axis of is inclined at an angle relative to the coordinate axis, where . We rotate the coordinate axis by angle anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let be the coordinates in the rotated system. Then and are related by In the rotated coordinate system, the parabola has focus at and the two pints on it are at and . Therefore, the directrix is ; we can, WLOG, choose . For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is From we have , so we need . Substituting in , we get For to be an integer must be a multiple of 5; setting we get Now we need to be odd, i.e. is an odd multiple of , in which case we get , which is also an integer. The values that satisfy the given conditions correspond to , and there are such numbers.
(Solution by Shaddoll)
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2014amc12a/384
~ dolphin7
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
1) The line of symmetry is NOT y= -x but 4x + 3y = 0
2) In the expression for x, it is NOT 8 but 8k.
With these minor corrections, the solution still holds good.